Deflection of beams: double integration and Macaulay's method

Why it matters

Stiffness often matters as much as strength. A wing that bends too far changes its aerodynamic loads and can flutter; a control-rod support that sags misrigs the controls; a landing-gear beam that deflects too much fouls nearby structure. The elastic-curve equation turns the bending moment you already know how to find into slopes and deflections, and Macaulay's method makes it practical for real load cases.

Key ideas

The elastic curve. From bending theory, curvature 1/R = M/(E·I). For small slopes, 1/R ≈ d²y/dx², giving the governing equation E·I·d²y/dx² = M(x). Here y is the deflection (taken positive upward in this lesson), x is measured along the beam and M is positive for sagging. With this convention a sagging moment gives positive curvature, and a downward-loaded beam has negative y.

Assumptions: linear elastic material, small deflections and slopes (so (dy/dx)² ≪ 1), plane sections remain plane, and deflection due to shear is neglected (fine for slender beams, span/depth above about 10).

Double integration. Write M(x), integrate once to get E·I·dy/dx (slope) plus a constant C₁, and again to get E·I·y plus C₂. The constants come from boundary conditions:

• Simple support or roller: y = 0.
• Fixed end: y = 0 and dy/dx = 0.
• Free end: no condition on y or slope (M and V are zero there).
• Symmetry: dy/dx = 0 at the line of symmetry. Maximum deflection occurs where dy/dx = 0 (or at a free end).

Why Macaulay's method. With several point loads, M(x) has a different expression in every segment, and ordinary double integration produces two constants per segment, all to be matched by continuity. Macaulay writes one moment expression for the whole beam using brackets ⟨x − a⟩ that are taken as zero when x < a. Integrate the brackets as a whole: ∫⟨x − a⟩ⁿ dx = ⟨x − a⟩ⁿ⁺¹/(n + 1). Only two constants appear, and continuity of slope and deflection is automatic.

Rules for Macaulay's method:

• Measure x from the left end and write M for a section in the last segment, taking all loads to the left.
• A point load W at a: term −W·⟨x − a⟩.
• A couple M₀ at a (clockwise): term +M₀·⟨x − a⟩⁰.
• A UDL that stops before the right end must be extended to the end and an equal upward UDL added to cancel the extension.
• The method needs a single E·I. For a stepped beam, use M/(E·I) diagrams with the moment-area method instead.

Standard results (memorise; they also check longer work):

• Cantilever, end load W: y_max = W·L³/(3EI), slope W·L²/(2EI).
• Cantilever, full UDL w: y_max = w·L⁴/(8EI), slope w·L³/(6EI).
• Simply supported, central load W: y_max = W·L³/(48EI), end slope W·L²/(16EI).
• Simply supported, full UDL w: y_max = 5·w·L⁴/(384EI), end slope w·L³/(24EI).

Formulas

E·I·d²y/dx² = M(x) E = Young's modulus (Pa), I = second moment of area (m⁴), y = deflection (m), x = position (m), M = bending moment (N·m), sagging positive.

⟨x − a⟩ⁿ = 0 for x < a; (x − a)ⁿ for x ≥ a Macaulay bracket; integrate as ⟨x − a⟩ⁿ⁺¹ / (n + 1).

δ = W·L³ / (3·E·I), θ = W·L² / (2·E·I) Cantilever, end load W (N), span L (m).

δ = w·L⁴ / (8·E·I), θ = w·L³ / (6·E·I) Cantilever, UDL w (N/m).

δ = W·L³ / (48·E·I), θ_end = W·L² / (16·E·I) Simply supported, central load.

δ = 5·w·L⁴ / (384·E·I), θ_end = w·L³ / (24·E·I) Simply supported, full UDL.

δ_load = W·a²·b² / (3·E·I·L) Simply supported, load W at distances a and b from the supports; deflection under the load.

Worked examples

Example 1 (standard). A simply supported beam of span 4 m carries a central load of 40 kN. E = 200 GPa and I = 8 × 10⁷ mm⁴ (8 × 10⁻⁵ m⁴). Derive and evaluate the midspan deflection and the end slope.

1. For 0 ≤ x ≤ L/2: M = (W/2)·x, so E·I·y'' = W·x/2.
2. Integrate: E·I·y' = W·x²/4 + C₁. Symmetry gives y' = 0 at x = L/2, so C₁ = −W·L²/16.
3. Integrate: E·I·y = W·x³/12 − W·L²·x/16 + C₂. y = 0 at x = 0 gives C₂ = 0.
4. At x = L/2: E·I·y = W·L³/96 − W·L³/32 = −W·L³/48.
5. E·I = 200 × 10⁹ × 8 × 10⁻⁵ = 1.6 × 10⁷ N·m².
6. δ = 40 000 × 4³/(48 × 1.6 × 10⁷) = 3.33 × 10⁻³ m.
7. End slope: θ = W·L²/(16EI) = 40 000 × 16/(16 × 1.6 × 10⁷) = 2.5 × 10⁻³ rad.

Answer: δ_mid ≈ 3.33 mm downward, θ_end = 0.0025 rad.

Example 2 (GATE level, Macaulay). A simply supported beam AB of span 6 m carries 30 kN at 2 m and 20 kN at 4 m from A. E·I = 2 × 10⁷ N·m². Find the deflection at midspan.

1. Reactions: R_A = (30 × 4 + 20 × 2)/6 = 26.67 kN; R_B = 50 − 26.67 = 23.33 kN.
2. E·I·y'' = 26 667·x − 30 000·⟨x − 2⟩ − 20 000·⟨x − 4⟩ (N·m).
3. E·I·y' = 13 333·x² − 15 000·⟨x − 2⟩² − 10 000·⟨x − 4⟩² + C₁.
4. E·I·y = 4444.4·x³ − 5000·⟨x − 2⟩³ − 3333.3·⟨x − 4⟩³ + C₁·x + C₂.
5. y = 0 at x = 0 gives C₂ = 0 (all brackets vanish).
6. y = 0 at x = 6: 4444.4 × 216 − 5000 × 64 − 3333.3 × 8 + 6·C₁ = 0, i.e. 960 000 − 320 000 − 26 667 + 6·C₁ = 0, so C₁ = −102 222 N·m².
7. At x = 3 m (the ⟨x − 4⟩ term is zero): E·I·y = 4444.4 × 27 − 5000 × 1 − 102 222 × 3 = 120 000 − 5000 − 306 667 = −191 667 N·m³.
8. y = −191 667/(2 × 10⁷) = −9.58 × 10⁻³ m.

Answer: 9.58 mm downward at midspan. (The true maximum, where y' = 0, is 9.59 mm at about 2.94 m, almost the same.)

Common mistakes

• Expanding a Macaulay bracket before integrating; integrate ⟨x − a⟩ as a single variable.
• Keeping a bracket term when evaluating at x < a; it must be dropped.
• Applying a UDL that stops short without the compensating opposite load.
• Unit slips with I: 1 cm⁴ = 10⁻⁸ m⁴ and 1 mm⁴ = 10⁻¹² m⁴.
• Using W (the total load) in place of w (load per metre) in 5wL⁴/(384EI).
• Using Macaulay with a changing E·I along the span.

For GATE AE

Expect direct use of the standard deflection and slope results, ratio questions (for example how deflection changes when span or depth doubles), the location of maximum deflection, and Macaulay problems with two point loads or a partial UDL. Practise writing the single Macaulay moment equation quickly and evaluating the constants from the two support conditions.

Quick check

1. What are the boundary conditions at a fixed end?
2. If the span of a simply supported beam under a UDL doubles, by what factor does y_max rise?
3. If the depth of a rectangular beam doubles, by what factor does its deflection fall?
4. Convert 6000 cm⁴ to m⁴.

Answers: 1. y = 0 and dy/dx = 0. 2. 16. 3. 8. 4. 6 × 10⁻⁵ m⁴.